Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int M[11000][34];
int Y[11000][34];
int x[11000], y[11000];
int n;
int um[11000];
vector<pair<int, int> > I, F;
int S(int a, int b) {
  I.push_back(pair<int, int>(a, b));
  I.push_back(pair<int, int>(b, a));
  I.push_back(pair<int, int>(a, b));
}
int PL(int a, int b) { I.push_back(pair<int, int>(a, b)); }
int SY(int a, int b) {
  F.push_back(pair<int, int>(a, b));
  F.push_back(pair<int, int>(b, a));
  F.push_back(pair<int, int>(a, b));
}
int PLY(int a, int b) { F.push_back(pair<int, int>(a, b)); }
int B() {
  cout << -1 << endl;
  exit(0);
}
int eh[11000];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> x[i];
  for (int i = 0; i < n; i++) cin >> y[i];
  for (int i = 0; i < n; i++) {
    int a = y[i];
    int topo = 34 - 1;
    while (a) {
      Y[i][topo--] = a % 2;
      a /= 2;
    }
  }
  for (int i = 0; i < n; i++) {
    int a = x[i];
    int topo = 34 - 1;
    while (a) {
      M[i][topo--] = a % 2;
      a /= 2;
    }
  }
  memset(um, -1, sizeof um);
  int c = 0;
  for (int i = 0; i < min(n, 34); i++) {
    if (c == 34) break;
    if (M[i][c] == 0) {
      int qm = -1;
      for (int j = i + 1; j < n; j++) {
        if (M[j][c]) qm = j;
      }
      if (qm == -1) {
        i--;
        c++;
        continue;
      }
      S(i, qm);
      for (int j = 0; j < 34; j++) swap(M[i][j], M[qm][j]);
    }
    um[c] = i;
    eh[i] = 1;
    for (int j = 0; j < n; j++)
      if (j != i && M[j][c]) {
        PL(i, j);
        for (int k = 0; k < 34; k++) M[j][k] ^= M[i][k];
      }
  }
  c = 0;
  for (int i = 0; i < min(n, 34); i++) {
    if (c == 34) break;
    if (Y[i][c] == 0) {
      int qm = -1;
      for (int j = i + 1; j < n; j++) {
        if (Y[j][c]) qm = j;
      }
      if (qm == -1) {
        i--;
        c++;
        continue;
      }
      SY(i, qm);
      for (int j = 0; j < 34; j++) swap(Y[i][j], Y[qm][j]);
    }
    for (int j = 0; j < n; j++)
      if (j != i && Y[j][c]) {
        PLY(i, j);
        for (int k = 0; k < 34; k++) Y[j][k] ^= Y[i][k];
      }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 34; j++) {
      if (M[i][j] != Y[i][j]) {
        if (um[j] != -1) {
          PL(um[j], i);
          for (int k = 0; k < 34; k++) M[i][k] ^= M[um[j]][k];
        }
      }
    }
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 34; j++)
      if (M[i][j] != Y[i][j]) B();
  printf("%d\n", I.size() + F.size());
  for (int i = 0; i < I.size(); i++)
    printf("%d %d\n", I[i].second + 1, I[i].first + 1);
  reverse(F.begin(), F.end());
  for (int i = 0; i < F.size(); i++)
    printf("%d %d\n", F[i].second + 1, F[i].first + 1);
  return 0;
}